Question

Calcular las siguientes potencias
A .2^0,27
B.3^-1,51
C.(1/3)^2,47
D.(1/2)^-2,05

Answer 1

Potencias: calculadas con el método Newton-Rhapson

A = 1.205

B = 1.200

C = 0.0621

D = 4.14

Para resolver las potencias, aproximamos el resultado mediante el método de Newton-Raphson,

definimos una función f de la siguiente manera

$f_t(x)= x^{1/t} - a$

Donde a es la base y t el exponente. El algoritmo de Newton nos dice que el resultado se puede aproximar de la siguiente manera:

$x_{m+1} = x_{m}[1 + t(\frac{a}{x_{m}^{1/t}} - 1) ]$

los casos iniciales se escogen de manera aproximada

1. $t = 0.27\\1/t = 3.7037\\a = 2\\x_0 = 1\\\\x_1 = 1 +0.27(\frac{2}{1}-1) = 1.27\\x_2 = 1.27[1+0.27(\frac{2}{1.27^{3.7037}}-1)] = 1.27[1-0.27*1747]= 1.21\\x_3 = 1.21 [1+0.27( \frac{2}{1.21^{3.7037}} -1)] = 1.21[1-0.27*0.0127]= 1.205$
2. $t = 1.51\\1/t = 0.6622\\a = 1/3\\x_0 = 0.5\\\\x_1 = 0.5[ 1 +1.51(\frac{1}{3*0.5^{0.6622}}-1) ] = 0.5(1-1.51*0.4725) = 0.1433\\\\x_2 = 0.1433[1+1.51( \frac{1}{3*0.1433^{0.6622}}-1 )] = 0.1433( 1 +1.51*0.2067) = 0.1880\\\\x_3 = 0.1880[ 1+1.51( \frac{1}{3*0.1880^{0.6622}}-1 ) ] = 0.1880( 1+1.51*0.0081) = 1.200$ Aquí hago uso de la propiedad a^(-t) = (1/a)^t
3. $t =2.47\\1/t = 0.4048\\a = 1/3\\x_0 = 0.1\\\\x_1 = 0.1[ 1+2.47( \frac{1}{3*0.1^{0.4048}}-1)] = 0.1[1-2.47*0.1533] =0.0621$
4. $t = 2.05\\1/t = 0.4878\\a = 2\\x_0 = 4\\\\x_1 = 4*(1+2.05*( \frac{2}{4^{0.4878}} -1)) = 4*(1+2.05*0.017) = 4.1394\\\\x_2 = 4.1394[1+2.05( \frac{2}{4.1394^{0.4878}} -1)] = 4.1394*(1+2.05*0.0002) = 4.141$